Is this correct? A local maximum for f(x) = xe^2x occurs where x = 2 .?

Question

A local maximum for f(x) = xe^2x
occurs where x = 2 .

Answer 1

No, x = 2 is not a local max.
f'(x) = x (e^2x)' + (x)' e^2x
f'(x) = x(2e^2x) + (1)e^2x
f'(x) = 2xe^2x + e^2x
2xe^2x + e^2x = 0
e^2x(2x + 1) = 0
e^2x = 0 or 2x + 1 = 0
e^2x never equals 0, so x = -1/2 is the only critical point.
It is a local min.

Answer 2

Take the first derivative and set equal to zero:

f'(x) = (x)(2e^2x) + (1)(e^2x) = 0

(e^2x)(2x + 1) = 0, x = -(1/2)

Take second derivative to find if this is a max or a min:

f''(x) = (x)(4e^2x) + (1)(2e^2x) +(2e^2x)
=(x)(4e^2x) + (4e^2x)
=(4e^2x)(x + 1)
At x = -(1/2)
f''(-1/2) = (4/e)(1/2)

Since f''(-1/2) > 0 you have x = -1/2 as a minimum.

Answer 3

this one requires integration. apply the mean value theorem, then integrate and solve for zero. I believe the max is close to two, within some abitrary epsilon.